Decagonal number

In mathematics, a decagonal number is a figurate number that extends the concept of triangular and square numbers to the decagon (a ten-sided polygon). However, unlike the triangular and square numbers, the patterns involved in the construction of decagonal numbers are not rotationally symmetrical. Specifically, the n-th decagonal numbers counts the dots in a pattern of n nested decagons, all sharing a common corner, where the ith decagon in the pattern has sides made of i dots spaced one unit apart from each other. The n-th decagonal number is given by the following formula

: $d_n = 4n^2 - 3n$ .

The first few decagonal numbers are:

: 0, 1, 10, 27, 52, 85, 126, 175, 232, 297, 370, 451, 540, 637, 742, 855, 976, 1105, 1242, 1387, 1540, 1701, 1870, 2047, 2232, 2425, 2626, 2835, 3052, 3277, 3510, 3751, 4000, 4257, 4522, 4795, 5076, 5365, 5662, 5967, 6280, 6601, 6930, 7267, 7612, 7965, 8326 {{OEIS|id=A001107}}.

The nth decagonal number can also be calculated by adding the square of n to thrice the (n−1)th pronic number or, to put it algebraically, as

: $D_n = n^2 + 3\left(n^2 - n\right)$ .

Properties

Decagonal numbers consistently alternate parity.
$D_n$ is the sum of the first $n$ natural numbers congruent to 1 mod 8.
$D_n$ is number of divisors of $48^{n-1}$ .
The only decagonal numbers that are square numbers are 0 and 1.
The decagonal numbers follow the following recurrence relations:

: $D_n=D_{n-1}+8n-7 , D_0=0$

: $D_n=2D_{n-1}-D_{n-2}+8, D_0=0,D_1=1$

: $D_n=3D_{n-1}-3D_{n-2}+D_{n-3}, D_0=0, D_1=1, D_2=10$

Sum of reciprocals

The sum of the reciprocals of the decagonal numbers admits a simple closed form:

$\sum_{n=1}^{\infty}\frac{1}{4n^{2}-3n}+\sum_{n=1}^{\infty}\frac{1}{n\left(4n-3\right)}=\ln\left(2\right)+\frac{\pi}{6}.$

=Proof=

This derivation rests upon the method of adding a "constructive zero":

$\begin{align}
\sum_{n=1}^{\infty}\frac{1}{n\left(4n-3\right)} & {} =\frac{4}{3}\sum_{n=1}^{\infty}\left(\frac{1}{4n-3}-\frac{1}{4n}\right) \\
&=\frac{2}{3}\sum_{n=1}^{\infty}\left(\frac{2}{4n-3}-\frac{2}{4n}+\left(\frac{1}{4n-1}-\frac{1}{4n-2}\right)-\left(\frac{1}{4n-1}-\frac{1}{4n-2}\right)\right)
\end{align}$

Rearranging and considering the individual sums:

$\begin{align}
&= \frac{2}{3} \sum_{n=1}^{\infty} \left[ \left(\frac{1}{4n-3} - \frac{1}{4n-2} + \frac{1}{4n-1} - \frac{1}{4n} \right) + \left(\frac{1}{4n-2} - \frac{1}{4n} \right) + \left(\frac{1}{4n-3} - \frac{1}{4n-1} \right) \right] \\
&= \frac{2}{3} \sum_{n=1}^{\infty} \left( \frac{1}{4n-3} - \frac{1}{4n-2} + \frac{1}{4n-1} - \frac{1}{4n} \right) + \frac{1}{3} \sum_{n=1}^{\infty} \left( \frac{1}{2n-1} - \frac{1}{2n} \right) + \frac{2}{3} \sum_{n=1}^{\infty} \left( \frac{1}{2(2n-1)-1} - \frac{1}{2(2n)-1} \right) \\
&= \frac{2}{3} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} + \frac{1}{3} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} + \frac{2}{3} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n-1} \\
&= \ln\left(2\right)+\frac{\pi}{6}.
\end{align}$

Category:Figurate numbers